Q40 / 150 · MHT CET 2026 (16 April Shift 2) · Chemistry
What is action of $\text{CH}_3\text{ONa}$ on $\text{CH}_3\text{CH}_2\text{Br}$
$\text{CH}_3\text{ONa} + \text{CH}_3\text{CH}_2\text{Br} \rightarrow$ ?
$\text{CH}_3\text{ONa} + \text{CH}_3\text{CH}_2\text{Br} \rightarrow$ ?
- Correct answerAFormation of $\text{CH}_3\text{OCH}_2\text{CH}_3 + \text{NaBr}$
- BFormation of $\text{CH}_3\text{OH} + \text{NaBr}$
- CFormation of $\text{CH}_3\text{CH}_2\text{OH}$
- DFormation of $(\text{CH}_3)_2\text{O}$
Solution
The reaction between sodium methoxide ($\text{CH}_3\text{ONa}$) and ethyl bromide ($\text{CH}_3\text{CH}_2\text{Br}$) is an example of Williamson ether synthesis.
The methoxide ion ($\text{CH}_3\text{O}^{-}$) acts as a nucleophile and attacks the primary alkyl halide ($\text{CH}_3\text{CH}_2\text{Br}$) via an $S_N2$ mechanism.
The reaction proceeds as follows:
$\text{CH}_3\text{ONa} + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{OCH}_2\text{CH}_3 + \text{NaBr}$
The products formed are ethyl methyl ether and sodium bromide.
Answer: Formation of $\text{CH}_3\text{OCH}_2\text{CH}_3 + \text{NaBr}$
The methoxide ion ($\text{CH}_3\text{O}^{-}$) acts as a nucleophile and attacks the primary alkyl halide ($\text{CH}_3\text{CH}_2\text{Br}$) via an $S_N2$ mechanism.
The reaction proceeds as follows:
$\text{CH}_3\text{ONa} + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{OCH}_2\text{CH}_3 + \text{NaBr}$
The products formed are ethyl methyl ether and sodium bromide.
Answer: Formation of $\text{CH}_3\text{OCH}_2\text{CH}_3 + \text{NaBr}$
Hence, the correct answer is option: 'A'.